Let be $S_n$ a binomially distributed random variable and $p$ its corresponding probability. We define $P(\{S_n=k\})= b(k;n,p):={n\choose k} p^k(1-p)^{n-k}$. Here $n$ denotes the number of Bernoulli trials and $k$ the number of successes.
In the context of the local limit theorem and the weak law of large numbers we have the following statement in our lecture:
For each $a>0:$ $$ \lim\limits_{n\to\infty} P(|S_n-np|\leq a)\leq (2a+1)\lim\limits_{n\to\infty}\max\{P(S_n)\}=0, $$ because we know that $\lim\limits_{n\to\infty}\max\limits_{0\leq k\leq n} b(k;n,p)=0.$
I am not sure how to interpret the expression $\max\{P(S_n)\}$ and why this inequality holds at all?
I assume that it has something to do with the fact that the set $\max\{S_n\}$ contains much more elements than the set $\{|S_n-np|\leq a\}$.
Can someone explain it to me or derive the inequality?
EDIT:
Maybe this is the idea behind it:
For sake of simplicity I assume that all numbers are naturals, otherwise I would add some floor/ceil functions.
Let's choose some $n$ and fix it. Then, there exists a $1\leq k_0\leq n$ such that for all $k$ with $1\leq k\neq k_0\leq n$ we have $P(\{S_n=k_0\})\geq P(\{S_n=k\})$. So the set $\{S_n=k_0\}$ represents some kind of maximum which corresponds to the expression above $\max\limits_{0\leq k\leq n} b(k;n,p)$. To me it would make more sense if the expression $\max\{P(S_n)\}$ was replaced by $\{S_n=k_0\}$.
Next, we use $\{S_n=k_0\}$ to construct an upper bound. From $$ |S_n-np|\leq a\iff np-a\leq S_n\leq np+a $$ we sse tha there are at most $(2a+1)$-many natural numbers between $np-a $ and $np+a$. So $S_n$ can attain at most $(2a+1)$-many different values i.e.: \begin{align*} &S_n=np-a\\ &S_n=np-a+1\\ &\dots\\ &S_n=np+a. \end{align*} Each of those $(2a+1)$-many possibilities generates a set. As above mentioned $\{S_n=k_0\}$ is the set with the highest probability, so \begin{align*} &P(|S_n-np|\leq a)=P(np-a\leq S_n\leq np+a)\\ &=P(\{S_n=np-a\})+P(\{S_n=np-a+1\})+\dots+P(\{S_n=np+a\})\\ &\leq (2a+1)P(\{S_n=k_0\})=(2a+1)\max\limits_{0\leq k\leq n} b(k;n,p). \end{align*} Finally, letting $n\to\infty$ shows the statement.
Is this correct?